JEE MAIN - Mathematics (2005 - No. 35)
If $${z_1}$$ and $${z_2}$$ are two non-zero complex numbers such that $$\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$, then arg $${z_1}$$ - arg $${z_2}$$ is equal to :
$${\pi \over 2}\,$$
$$ - \pi $$
0
$${{ - \pi } \over 2}$$
Explanation
Given that, $$\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$
$$\,\left| {{z_1} + {z_2}} \right|$$ is the vector sum of $${z_1}$$ and $${z_2}$$. So $$\,\left| {{z_1} + {z_2}} \right|$$ should be $$<$$ $$\left| {{z_1}} \right| + \left| {{z_2}} \right|$$ but here they are equal so $${z_1}$$ and $${z_2}$$ are collinear.
S if $${z_1}$$ makes an angle $$\theta $$ with x axis then $${z_2}$$ will also make $$\theta $$ angle.
$$\therefore$$ arg $${z_1}$$ - arg $${z_2}$$ = $$\theta $$ - $$\theta $$ = 0
$$\,\left| {{z_1} + {z_2}} \right|$$ is the vector sum of $${z_1}$$ and $${z_2}$$. So $$\,\left| {{z_1} + {z_2}} \right|$$ should be $$<$$ $$\left| {{z_1}} \right| + \left| {{z_2}} \right|$$ but here they are equal so $${z_1}$$ and $${z_2}$$ are collinear.
S if $${z_1}$$ makes an angle $$\theta $$ with x axis then $${z_2}$$ will also make $$\theta $$ angle.
$$\therefore$$ arg $${z_1}$$ - arg $${z_2}$$ = $$\theta $$ - $$\theta $$ = 0
Comments (0)
